A solution of Fe2(SO4)3 is electrolyzed for ‘x’ min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is _______. [nearest integer]
Given : 1 F = 96500 C mol–1.
Atomic mass of Fe = 56 g mol–1
A solution of Fe2(SO4)3 is electrolyzed for ‘x’ min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is _______. [nearest integer]
Given : 1 F = 96500 C mol–1.
Atomic mass of Fe = 56 g mol–1
Correct Answer: 20
Solution and Explanation
Fe3+ + 3e– → Fe
Moles of Fe deposited
\(=\frac{0.3482}{56}=6.2×10^{-3}\)
For 1 mole Fe, charge required is 3 F
For 6.2 × 10–3 mole Fe, charge required is
3 × 6.2 × 10–3 F
Since, charge required = 18.6 × 10–3 × 96500 C
= 1794.9 C
And,
1.5 × t = 1794.9
\(t=\frac{1794.9}{1.5×60}\) min
t ≃20 min
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